What Are Imaginary Numbers?

What is \(\sqrt{-4}\)? If you try to answer using only the real numbers you've worked with before, you're stuck — no real number squared gives a negative result. Positive times positive is positive, negative times negative is also positive.

That's not a gap in your understanding. It's a genuine gap in the real number system. Imaginary numbers exist specifically to fill it.

Defining i

Mathematicians defined a new number, called \(i\), to represent the square root of \(-1\):

$$i = \sqrt{-1} \qquad \text{equivalently,} \qquad i^2 = -1$$

This is the definition. \(i\) is not a variable — it's a specific constant, just like \(\pi\). It doesn't live on the real number line, which is why it's called imaginary. But it obeys all the usual rules of algebra, which is what makes it useful.

With \(i\) defined, the square root of any negative number becomes available:

$$\sqrt{-4} = \sqrt{4 \cdot (-1)} = \sqrt{4} \cdot \sqrt{-1} = 2i$$

$$\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i$$

$$\sqrt{-25} = 5i$$

The pattern: factor out \(-1\), take the square root of the positive part normally, and attach \(i\).

The Powers of i

What happens when you raise \(i\) to higher powers? Something interesting:

Power	Calculation	Result
\(i^1\)	\(i\)	\(i\)
\(i^2\)	definition	\(-1\)
\(i^3\)	\(i^2 \cdot i = -1 \cdot i\)	\(-i\)
\(i^4\)	\(i^2 \cdot i^2 = (-1)(-1)\)	\(1\)
\(i^5\)	\(i^4 \cdot i = 1 \cdot i\)	\(i\)

The pattern repeats with period 4: \(i,\ -1,\ -i,\ 1,\ i,\ -1,\ -i,\ 1, \ldots\)

To find any power of \(i\), divide the exponent by 4 and look at the remainder:

• Remainder 0 → \(1\)
• Remainder 1 → \(i\)
• Remainder 2 → \(-1\)
• Remainder 3 → \(-i\)

Example: What is \(i^{23}\)?

\(23 \div 4 = 5\) remainder \(3\), so \(i^{23} = -i\).

Simplifying Square Roots of Negatives

To simplify \(\sqrt{-n}\) for any positive \(n\):

1. Factor: \(\sqrt{-n} = \sqrt{n} \cdot \sqrt{-1} = \sqrt{n} \cdot i\)
2. Simplify \(\sqrt{n}\) as far as possible

Example 1

Simplify \(\sqrt{-80}\).

Factor into a perfect square and the rest:

$$\sqrt{-80} = \sqrt{16 \cdot 5 \cdot (-1)} = \sqrt{16} \cdot \sqrt{5} \cdot \sqrt{-1} = 4\sqrt{5} \cdot i = 4i\sqrt{5}$$

Example 2

Simplify \((4i)^2\).

$$(4i)^2 = 16i^2 = 16(-1) = -16$$

Squaring \(4i\) gives a real (and negative) answer — this is the whole point of imaginary numbers.

Example 3

Simplify \(i^3 + i^4\).

$$i^3 + i^4 = -i + 1 = 1 - i$$

Complex Numbers

Imaginary numbers rarely travel alone. More often you'll see them combined with a real number in what's called a complex number:

$$a + bi$$

where \(a\) is the real part and \(b\) is the imaginary part. Both \(a\) and \(b\) are ordinary real numbers.

Examples:

• \(3 + 2i\) — real part 3, imaginary part 2
• \(-1 - 5i\) — real part −1, imaginary part −5
• \(7\) — a real number is just a complex number with \(b = 0\)
• \(4i\) — a pure imaginary number has \(a = 0\)

Complex numbers arise naturally when solving quadratic equations with negative discriminants. For example, \(x^2 + 1 = 0\) has no real solutions, but it has two complex solutions: \(x = i\) and \(x = -i\).

Adding, Subtracting, and Multiplying

Adding and subtracting: Combine real parts with real parts, imaginary parts with imaginary parts.

$$(3 + 2i) + (1 - 5i) = (3+1) + (2-5)i = 4 - 3i$$

$$(6 - i) - (2 + 3i) = (6-2) + (-1-3)i = 4 - 4i$$

Multiplying: Use FOIL, then replace \(i^2\) with \(-1\).

$$(3 + 2i)(1 - 4i) = 3(1) + 3(-4i) + 2i(1) + 2i(-4i)$$

$$= 3 - 12i + 2i - 8i^2$$

$$= 3 - 10i - 8(-1)$$

$$= 3 - 10i + 8 = 11 - 10i$$

The \(i^2 = -1\) substitution is what brings the result back to \(a + bi\) form.

Practice Problems

Simplify \(\sqrt{-36}\).

Show answer\(\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i\)

Simplify \((3i)^2\).

Show answer\((3i)^2 = 9i^2 = 9(-1) = -9\)

Simplify \(\sqrt{-48}\).

Show answer\(\sqrt{-48} = \sqrt{16 \cdot 3 \cdot (-1)} = 4\sqrt{3} \cdot i = 4i\sqrt{3}\)

What is \(i^{30}\)?

Show answer\(30 \div 4 = 7\) remainder \(2\), so \(i^{30} = i^2 = -1\).

Add: \((5 + 3i) + (2 - 7i)\).

Show answer\((5+2) + (3-7)i = 7 - 4i\)

Multiply: \((2 + i)(3 - 2i)\).

Show answerFOIL: \(6 - 4i + 3i - 2i^2 = 6 - i - 2(-1) = 6 - i + 2 = 8 - i\)